Question

The function$f\left(x\right)={(x–3)}^2$satisfies all the conditions of the mean value theorem in $[3,4].$ A point on $y={(x–3)}^2$, where the tangent is parallel to the chord joining$(3,0)$ and $(4,1)$ is

• A. $\left(\frac{7}{2},\frac{1}{2}\right)$
• B. $\left(\frac{7}{2},\frac{1}{4}\right)$
• C. $(1,4)$
• D. $(4,1)$

Answer 1

The correct option is B

$\left(\frac{7}{2},\frac{1}{4}\right)$

Evaluating the correct answer.

Let the point be $(x_1,y_1)$

So, the slope of the tangent is

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=\frac{d{(x-3)}^2}{dx} \\ \Rightarrow f\text{'}\left(x\right)=2(x-3) \end{gathered}$$

So, the slope of the chord is

$\Rightarrow \frac{(y_2-y_1)}{\left(x_2-x_1\right)}=\frac{\left(1-0\right)}{\left(4-3\right)}=1$

The slope of tangent $=$ slope of the chord [ if two lines are parallel then their slopes are equal ]

$$\begin{gathered} \Rightarrow 2(x_1-3)=1 \\ \Rightarrow x_1-3=\frac{1}{2} \\ \Rightarrow x_1=\frac{7}{2} \end{gathered}$$

$\therefore y={(x-3)}^2$

Putting the value of $x$ in the above equation

$$\begin{gathered} \Rightarrow y_1={\left(\frac{7}{2}-3\right)}^2 \\ \Rightarrow y_1=\left(\frac{1}{4}\right) \end{gathered}$$

Hence, the correct option is B.